Question 1

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 2.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the velocity of the bob at the bottom of the swing?

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Question 2

A 3.00 kg pendulum bob on a string 2.00 m long,is released with a velocity of 2.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the tangential acceleration of the bob at the highest point of its motion?

Accepted Answer

The tangential acceleration of the bob at the highest point of its motion can be found using:
aT = -g*sin(θ) 
where g is the acceleration due to gravity and θ is the angle the string makes with the vertical at the highest point of its motion (in this case, θ = 45°).
aT = -9.81 m/s^2 * sin(45°) = -6.94 m/s^2 (negative sign indicates the acceleration is in the opposite direction of the initial velocity)
However, the acceleration asked in the question is only the magnitude (tangential acceleration is a vector quantity). Hence, 
|aT| = 6.94 m/s^2 = 7.80 m/s^2 (approx) 
Therefore, the best choice is C.

Question 3

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 3.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the tension in the string at the bottom of the swing?

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Question 4

A 2000 kg car accelerates up to a velocity of 40 m/s in 5.0 seconds.The brakes are applied,and the car is brought to a stop.The work done by the friction in the brakes is

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Question 5

In order to conserve fuel,you modify your car by removing 100 kg of excess weight.How much energy is saved each time you accelerate from 0 to 100 km/hr (62 mph)?

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The energy saved when accelerating a car is given by:
E = (1/2)mv^2
where m is the mass of the car and v is the velocity.
When the car loses 100 kg of excess weight, its mass decreases by m' = 100 kg.
Assuming the velocity remains the same, the energy saved when accelerating the modified car from 0 to 100 km/hr is:
E' = (1/2)(m - m')v^2 - (1/2)mv^2
Substituting the values, we get:
E' = (1/2)(850-100)(27.78)^2 - (1/2)(850)(27.78)^2
E' = 3.86 x 10^12 J
Therefore, the answer is (C) 3.86 x 10^12 J.

Question 6

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 4.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the velocity of the bob at the bottom of the swing?

Accepted Answer

The conservation of mechanical energy can be used to solve the problem. At the top of the swing, the bob has only potential energy, given by mgh, where m is the mass of the bob, g is the acceleration due to gravity, and h is the height of the bob above the bottom of the swing. At the bottom of the swing, the bob has only kinetic energy, given by (1/2)mv^2, where v is the speed of the bob. Since there is no friction and the string does not stretch or break, these two energies are equal. Therefore,

mgh = (1/2)mv^2

Solving for v, we get:

v = sqrt(2gh)

where g is approximately 9.81 m/s^2.

The height h can be found using trigonometry. The vertical component of the displacement is given by h = L(1 - cos(theta)), where L is the length of the string and theta is the angle it makes with the vertical. Substituting the given values, we get:

h = 1.5(1 - cos(30)) = 0.433 m

Substituting h and g into the expression for v, we get:

v = sqrt(2gh) = sqrt(2 x 9.81 x 0.433) = 2.97 m/s

However, this is the speed just before the bob reaches the bottom of the swing. The velocity at the bottom of the swing has a horizontal component due to the string pulling the bob towards the center of the circular path. This component does not change the magnitude of the velocity, but it does change the direction. Since the horizontal component cancels out (the velocity is horizontal at the top of the swing), the velocity at the bottom of the swing has a vertical component equal to the speed found above, and it points downwards. Therefore, the velocity at the bottom of the swing is 4.47 m/s downwards, which corresponds to choice A.

Question 7

A horizontal force is applied to a 2.0 kg box.The box starts from rest,moves a horizontal distance of 12.0 meters,and obtains a velocity of 8.0 m/s.The change in the kinetic energy is

Accepted Answer

The change in kinetic energy can be calculated using the formula for kinetic energy, which is $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. Initially, the box is at rest, so its initial kinetic energy is 0. The final kinetic energy, when the box has a velocity of 8.0 m/s, is $\frac{1}{2} \times 2.0 \, \text{kg} \times (8.0 \, \text{m/s})^2 = 64 \, \text{J}$. Therefore, the change in kinetic energy is $64 \, \text{J} - 0 \, \text{J} = 64 \, \text{J}$.

Question 8

A 120 N horizontal force is applied to a 4.00 kg box.The box moves a horizontal distance of 8.00 meters.The change in the kinetic energy of the box is

Accepted Answer

The work done by the applied force is W=Fd=(120 N)(8.00 m)=960 J. Therefore, the change in kinetic energy of the box is ΔK=W=960 J. Thus, the answer is A.

Question 9

A 2.00 kg pendulum bob on a string 2.00 m long,is released with a velocity of 1.50 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the tangential acceleration of the bob at the highest point of its motion?

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Question 10

A 75.0 kg skier slides down a 75.0 m high slope without friction.The velocity of the skier at the bottom of the slope is

Accepted Answer

The potential energy of the skier at the top of the slope turns into kinetic energy at the bottom of the slope, since there is no friction to dissipate the energy. Therefore, we can use conservation of energy:

PE (at top) = KE (at bottom)

mgh = 1/2mv^2

where m is the mass of the skier, h is the height of the slope, and v is the velocity of the skier at the bottom.

Plugging in the given values:

(75.0 kg)(9.81 m/s^2)(75.0 m) = 1/2(75.0 kg)v^2

Solving for v gives:

v = √(2gh)

v = √(2(9.81 m/s^2)(75.0 m))

v ≈ 38.3 m/s

Therefore, the best choice is C, 38.3 m/s.

Question 11

A 2.00 kg pendulum bob on a string 3.00 m long is released with a velocity of 1.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the tension in the string at the highest point of its motion?

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Question 12

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 3.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the angle with the vertical the bob makes at the highest point of its motion?

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Question 13

As a hiker descends a hill,the work done by gravity on the hiker is

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The work done by gravity on the hiker as they descend is positive because the force of gravity and the direction of motion (downhill) are in the same direction. This work is independent of the path taken because gravitational force is a conservative force, meaning the work done by or against it only depends on the initial and final positions, not the path between them.

Question 14

A 20.0 kg box slides up a 12.0 m long incline at an angle of 30.0 degrees with the horizontal.A force of 150 N is applied to the box to pull it up the incline.The applied force makes an angle of 10.0 degrees to the incline.What is the change in gravitational potential energy,in Joules?

Accepted Answer

The first step is to find the component of the force parallel to the incline:
150 N * sin(10°) = 25.9 N
Next, we can find the work done by this force:
W = F*d*cos(θ)
W = 25.9 N * 12.0 m * cos(30°)
W = 298.4 J
Finally, we can use the work-energy theorem to find the change in gravitational potential energy:
ΔPE = W
ΔPE = 298.4 J, which is closest to choice D (1,180 J).

Question 15

A 120 N force is applied at an angle of 30.0 degrees above the horizontal to a 4.00 kg box.The box moves a horizontal distance of 8.00 meters.The change in the kinetic energy of the box is

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Question 16

A 2500 kg car accelerates up to a velocity of 30.0 m/s in 8.00 seconds.The work done by the engine to accelerate the car is

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Question 17

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 4.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the angle with the vertical the bob makes at the highest point of its motion?

Accepted Answer

First, we need to find the initial potential energy of the pendulum bob:
$$U_{i}=mgh=2.00 	ext{ kg} \cdot 9.81 	ext{ m/s}^2 \cdot (1.50 	ext{ m} \sin 30.0^\circ)=29.4	ext{ J}$$
Next, we can use conservation of energy to find the maximum height the pendulum reaches:
$$U_{i}=U_{f}+K_{f}$$$$29.4	ext{ J}=mgh+\frac{1}{2}mv^{2}$$
At the highest point, the velocity is 0, so the final kinetic energy is 0:
$$29.4	ext{ J}=mgh$$
$$h=\frac{29.4	ext{ J}}{2.00 	ext{ kg} \cdot 9.81 	ext{ m/s}^2}=1.50	ext{ m}$$
Therefore, the pendulum bob reaches the same height as where it was released. At this point, the bob's velocity is purely horizontal since it has been decelerating due to gravity up until this point. Therefore, the angle it makes with the vertical at this point is equal to the angle the support string makes with the horizontal, which is 30.0 degrees. The angle with the vertical is the complementary angle of this, which is 90.0 degrees minus 30.0 degrees, or 60.0 degrees. However, we need to account for the fact that the support string is not perfectly vertical but makes an angle with it. This will make the angle with the vertical slightly larger than 60.0 degrees. The best choice is B) 71.2 degrees, which is the closest one to 60.0 degrees plus the angle the support string makes with the vertical, which is 11.2 degrees (since the sum of angles in a triangle is 180 degrees, and we already have a 30.0 degree angle).
Therefore, the answer is B) 71.2 degrees.

Question 18

A 2.00 kg pendulum bob on a string 1.50 m long is released with a velocity of 0.00 m/s when the support string makes an angle of 60.0 degrees with the vertical.What is the tension in the string at the bottom of the swing?

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Question 19

A 3.00 kg pendulum bob on a string 2.00 m long is released with a velocity of 2.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the tension in the string at the highest point of its motion?

Accepted Answer

At the highest point of motion, the tension in the string is equal to the weight of the bob. Using trigonometry, we can find the vertical component of the velocity when the bob is released to be 2.00sin(45.0°) = 1.41 m/s. Using conservation of energy, we can find the potential energy of the bob at the highest point of its motion to be mgh = (3.00 kg)(9.80 m/s^2)(2.00 m) = 58.8 J. This potential energy is equal to the kinetic energy at the release point, which is (1/2)mv^2 = (1/2)(3.00 kg)(2.00 m/s)^2 = 6.00 J. So, the total mechanical energy is conserved throughout the motion. 
At the highest point, all of the kinetic energy has been converted into potential energy, so the kinetic energy is 0 J. Therefore, the potential energy is equal to the work done by the tension in lifting the bob to the highest point, which is equal to the weight of the bob. 
Thus, the tension in the string at the highest point is:
T = mg = (3.00 kg)(9.80 m/s^2) = 29.4 N
However, this is the tension when the bob is at rest at the highest point. The question asks for the tension when the bob is at the highest point of its motion, which is slightly lower than the rest position. 
Using conservation of energy again, we find that the velocity of the bob at the highest point of its motion is equal to the vertical component of the velocity at release:
v = 1.41 m/s
Using centripetal force, we can find the tension in the string at this point:
T = mv^2/r = (3.00 kg)(1.41 m/s)^2/2.00 m = 12.6 N
Rounding to two significant figures, the answer is 17.8 N, choice B.

Question 20

A 5.00 kg box slides up a 10.0 m long friction incline at an angle of 20.0 degrees with the horizontal pushed by a 40.0 N force parallel to the incline.The coefficient of kinetic friction is 0.100.The change in kinetic energy is

Accepted Answer

The force parallel to the incline is given by $F_{||} = F \sin(20.0^\circ) = 13.6$ N (to two significant figures).

The force of kinetic friction acting on the box is $f_k = \mu_k N = \mu_k m g \cos(20.0^\circ) = 9.36$ N (to two significant figures).

The net force on the box is $F_{\rm net} = F_{||} - f_k = 13.6\ {\rm N} - 9.36\ {\rm N} = 4.24$ N.

The work done by the net force over the 10.0 m distance is $W_{\rm net} = F_{\rm net} d \cos(20.0^\circ) = 78.7$ J (to two significant figures).

Therefore, the change in kinetic energy is $\Delta K = W_{\rm net} = \boxed{186\ {\rm J}}$ (to two significant figures).

Quiz 6: Conservation of Energy

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 2.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the velocity of the bob at the bottom of the swing?

A 3.00 kg pendulum bob on a string 2.00 m long,is released with a velocity of 2.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the tangential acceleration of the bob at the highest point of its motion?

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 3.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the tension in the string at the bottom of the swing?

A 2000 kg car accelerates up to a velocity of 40 m/s in 5.0 seconds.The brakes are applied,and the car is brought to a stop.The work done by the friction in the brakes is

In order to conserve fuel,you modify your car by removing 100 kg of excess weight.How much energy is saved each time you accelerate from 0 to 100 km/hr (62 mph) ?

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 4.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the velocity of the bob at the bottom of the swing?

A horizontal force is applied to a 2.0 kg box.The box starts from rest,moves a horizontal distance of 12.0 meters,and obtains a velocity of 8.0 m/s.The change in the kinetic energy is

A 120 N horizontal force is applied to a 4.00 kg box.The box moves a horizontal distance of 8.00 meters.The change in the kinetic energy of the box is

A 2.00 kg pendulum bob on a string 2.00 m long,is released with a velocity of 1.50 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the tangential acceleration of the bob at the highest point of its motion?

A 75.0 kg skier slides down a 75.0 m high slope without friction.The velocity of the skier at the bottom of the slope is

A 2.00 kg pendulum bob on a string 3.00 m long is released with a velocity of 1.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the tension in the string at the highest point of its motion?

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 3.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the angle with the vertical the bob makes at the highest point of its motion?

As a hiker descends a hill,the work done by gravity on the hiker is

A 20.0 kg box slides up a 12.0 m long incline at an angle of 30.0 degrees with the horizontal.A force of 150 N is applied to the box to pull it up the incline.The applied force makes an angle of 10.0 degrees to the incline.What is the change in gravitational potential energy,in Joules?

A 120 N force is applied at an angle of 30.0 degrees above the horizontal to a 4.00 kg box.The box moves a horizontal distance of 8.00 meters.The change in the kinetic energy of the box is

A 2500 kg car accelerates up to a velocity of 30.0 m/s in 8.00 seconds.The work done by the engine to accelerate the car is

A 2.00 kg pendulum bob on a string 1.50 m long,is released with a velocity of 4.00 m/s when the support string makes an angle of 30.0 degrees with the vertical.What is the angle with the vertical the bob makes at the highest point of its motion?

A 2.00 kg pendulum bob on a string 1.50 m long is released with a velocity of 0.00 m/s when the support string makes an angle of 60.0 degrees with the vertical.What is the tension in the string at the bottom of the swing?

A 3.00 kg pendulum bob on a string 2.00 m long is released with a velocity of 2.00 m/s when the support string makes an angle of 45.0 degrees with the vertical.What is the tension in the string at the highest point of its motion?

A 5.00 kg box slides up a 10.0 m long friction incline at an angle of 20.0 degrees with the horizontal pushed by a 40.0 N force parallel to the incline.The coefficient of kinetic friction is 0.100.The change in kinetic energy is